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Energy Flows and Balances. Units of Measure. BTU – amount of energy required to heat one pound of water, one degree Fahrenheit. Calorie – amount of energy required to heat 1 ml water 1 degree Celsius. Energy Balances and Conversion.

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Presentation Transcript
slide2

Units of Measure

BTU – amount of energy required to heat one pound of

water, one degree Fahrenheit

Calorie – amount of energy required to heat 1 ml

water 1 degree Celsius

slide3

Energy Balances and Conversion

The form of the available energy is often not the form that

is the most useful so it is common to have to convert one

form of energy to another.

Water in lake – turbine – electricity – light and heat

Bio-mass – combustion – steam - generator - electricity

Wind – windmill – generator - electricity

Conversion always less than 100% efficient

slide4

0

Rate of

Energy

accumulated

Rate of

Energy

In

Rate of

Energy

Out

Rate of

Energy

Generated

=

-

+

Rate of

Energy

Out

Rate of

Energy

Wasted

Rate of

Energy

In

Useful energy Out

=

-

X 100

Energy In

Since energy “flows” can use the same concepts as

materials balance to analyze

At Steady-State

Efficiency:

slide6

Calorimeter Example

A calorimeter holds 4 liters of water. When a 10 gram sample of a

waste-derived fuel is combusted the result is a 12.5o C rise in temperature.

What is the energy value of the fuel?

Energy In = Energy Out

(The idea behind a calorimeter is that no energy is wasted. It is all captured in the device.)

Energy Out = 12.5o C x 4 L x 103 ml/L x 1 g/ml

= 50 x 103(0C). G, or calories

= 50 x 103 calories x 4.18 (J/cal) = 209 x 103 J

Energy In = 209 x 103 J

Energy Value of the Fuel = (209 x 103 J/g)/ (10 g) = 20,900 J/g

slide7

Heat

Energy

Mass of

Material

Absolute Temperature

Of the Material

=

X

Heat Energy

This is only true when the heat capacity of the material is

independent of temperature. In particular when a phase

change occurs this is not true. (Water to Steam)

Energy Balance at Steady-State with two inflows

0 = Heat Energy In - Heat Energy Out + 0

0 = [T1Q1 + T2Q2] - T3Q3

T3 = [T1Q1 + T2Q2] / Q3

Also: Q3 = Q1 + Q2

slide8

Example

A coal-fired power plant discharges 3 m3/sec of cooling water at 80o C

into a river that has a flow of 15 m3/sec and a temperature of 20o C.

What will the temperature of the river be immediately downstream of

the discharge?

T3 = [T1Q1 + T2Q2] / Q3

= [(80 + 273)(3) + (20 +273)(15)] / (3 + 15)

= 303oK = 30o C

slide9

Energy Sources and Availability

Non-Renewable Sources

Nuclear Power

Coal, Peat, and Similar Products

Oil

Natural Gas

Renewable Sources

Hydropower from Rivers

Hydropower from Tides

Wood and Other Bio-mass

Solar Power

Refuse and other Wastes

Wind

slide11

Energy Equivalence

Arithmetic Energy Equivalence – based on energy amounts only

Conversion Energy Equivalence – takes into account the energy loss in

conversion

For Example

If gasoline has an energy value of 20,000 BTU/lb and refuse-

derived fuel has an energy value of 5,000 BTU/lb, the

arithmetic energy equivalence is:

20,000/5,000 = 4 lb refuse / 1 lb gasoline

It has been estimated that 50% of the energy in refuse derived

fuel is required for processing, therefore, the actual net energy

in the refuse is 2,500 BTU/lb. So:

Conversion energy equivalence = 20,000 / 2,500 = 8 lb refuse / 1 lb gasoline

slide12

Electric Power Production

Present power plans are less than 40% efficient

slide13

0, S.S.

Rate of

Energy

accumulated

Rate of

Energy

In

Rate of

Useful

Energy Out

Rate of

Wasted

Energy Out

=

-

-

Simplified:

Heat Engine

Energy Balance

0 = Qo - QU - QW

Efficiency (%) = QU/Qo

slide14

The most efficient engine possible is called a Carnot Engine. Its efficiency

is calculated as:

EC(%) = (T1 – To)/T1 x 100

Where: T1 = absolute temperature of the boiler

T2 = absolute temperature of the cooling water

Since this is the best possible:

(QU/Qo) < (T1 – To)/T1

Typical conditions for a power plant are: T1 = 600 + 273 = 873, and

T0 = 20 + 273 = 293

EC = (873 – 293) / 873 = 66%

Because real power plants have many other types of energy losses

(heat in stack gases, evaporation, friction) their actual efficiency is

about 40%. This figure is confirmed by operational data.

slide15

Where does all of this energy go?

All of it is dissipated in some way into the environment

60% of the energy content of the fuel that comes into

the plant is released to the environment as heat

15% stack gases, 45% cooling water

Thermal pollution

slide16

Cooling this water before discharge is a significant problem

Cooling towers such as these can add

up to 250% to the cost of a nuclear

power plant

Why is it better to allow this heat to be discharged to the

atmosphere rather than to a water body?

What else could you do with this heat?